Hi, I'm trying to call a script with the "?=" after the php script name in the URL, but can't seem to pick up the variable to use in my script... any help would be greatly appreciated :) I'm calling the script with this url: http://localhost/Data/test/dbconnect2c.php?id=TestPage The section of the script i'm using is: $id = $_GET['id']; $sql="SELECT * FROM content where title=$id"; $result=mysql_query($sql,$db); Error message returned: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in... Thanks, Gav This e-mail and any attachments are intended solely for the named addressee, are confidential and may contain legally privileged information. The copying or distribution of them or of any information they contain, by anyone other than the addressee, is prohibited. If you received this e-mail in error, please notify us immediately by return e-mail or telephone +61 2 9413 2944 and destroy the original message. Thank you. As Email is subject to viruses we advise that all Emails and any attachments should be scanned by an up to-date Anti Virus programme automatically by your system. It is the responsibility of the recipient to ensure that all Emails and any attachments are cleared of Viruses before opening. KSG can not accept any responsibility for viruses that maybe contained here in. Please advise KSG by return Email if you believe any Email sent by our system may contain a virus. It should be noted that most Anti Virus programmes can not scan encrypted file attachments (example - documents saved with a password). Thus extra care should be taken when opening these files. Liability limited by the Accountants Scheme, approved under the Professional Standards Act 1994 (NSW). Level 4 54 Neridah Street PO Box 1290 CHATSWOOD NSW 2067 CHATSWOOD NSW 2057 Ph: +61 2 9413 2944 Fax: +61 2 9413 9901 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
